A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of $100 \mathrm{~V} / \mathrm{m}$ makes it to move in a straight path, then the mass of the particle is (given charge of electron $=1.6 \times 10^{-19} \mathrm{C}$)

(D) $2.0 \times 10^{-24} \mathrm{~kg}$

(A) $9.1 \times 10^{-31} \mathrm{~kg}$

(B) $1.6 \times 10^{-27} \mathrm{~kg}$

(C) $1.6 \times 10^{-19} \mathrm{~kg}$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Magnetic Effects of Current

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of

100 V / m

makes it to move in a straight path, then the mass of the particle is (given charge of electron

= 1.6 \times 1 0^{- 19} C

)

(A) $9.1 \times 1 0^{- 31} kg$
(B) $1.6 \times 1 0^{- 27} kg$
(C) $1.6 \times 1 0^{- 19} kg$
(D) $2.0 \times 1 0^{- 24} kg$ ✓ Correct

Answer: (D)

Solution

This is a velocity-selector idea: the particle travels straight only when the electric and magnetic forces cancel,

q E = q v B

, so

v = \frac{E}{B}

. But before the field was added it was bent into a circle of radius

r

by the magnetic force alone, where

r = \frac{m v}{q B}

. Eliminating

v

m = \frac{q B r}{v} = \frac{q B ^{2} r}{E} = \frac{( 1.6 \times 1 0 ^{- 19} ) ( 0.5 ) ^{2} ( 0.5 \times 1 0 ^{- 2} )}{100} = 2.0 \times 1 0^{- 24} kg

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Magnetic Effects of Current

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